Key Equations

Medical Physics | AQA A-Level Physics

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Thin lens equation
$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$
  • Where:
    • $f$ = focal length (m)
    • $u$ = object distance (m)
    • $v$ = image distance (m)
Real-is-positive sign convention: positive $v$ for real images, negative $v$ for virtual images. Positive $f$ for converging lenses, negative $f$ for diverging lenses.
Half-value thickness
$$x_{1/2} = \frac{\ln 2}{\mu}$$
  • Where:
    • $x_{1/2}$ = half-value thickness (m)
    • $\mu$ = linear attenuation coefficient (m$^{-1}$)
The thickness of absorber needed to reduce the X-ray intensity to half its initial value. Derived by setting $I = I_0 / 2$ in the attenuation equation and solving for $x$.
Intensity level (decibel scale)
$$\text{IL} = 10 \log_{10}\left(\frac{I}{I_0}\right)$$
  • Where:
    • $\text{IL}$ = intensity level (dB)
    • $I$ = intensity of the sound (W m$^{-2}$)
    • $I_0$ = threshold of hearing ($1 \times 10^{-12}$ W m$^{-2}$)
Every increase of 10 dB corresponds to a tenfold increase in intensity. A doubling of intensity gives an increase of about 3 dB.
Intensity of sound
$$I = \frac{P}{A}$$
  • Where:
    • $I$ = intensity (W m$^{-2}$)
    • $P$ = power of the sound source (W)
    • $A$ = area over which the sound is spread (m$^2$)
For a point source radiating uniformly, $A = 4\pi r^2$, so intensity follows an inverse square law with distance.
X-ray attenuation
$$I = I_0\, e^{-\mu x}$$
  • Where:
    • $I$ = transmitted intensity (W m$^{-2}$)
    • $I_0$ = incident intensity (W m$^{-2}$)
    • $\mu$ = linear attenuation coefficient (m$^{-1}$)
    • $x$ = thickness of absorber (m)
Exponential decay of X-ray intensity with thickness. $\mu$ depends on the photon energy and the absorbing material. Higher $\mu$ means greater absorption.
Critical angle (fibre optics)
$$\sin \theta_c = \frac{n_2}{n_1}$$
  • Where:
    • $\theta_c$ = critical angle
    • $n_1$ = refractive index of the core (denser medium)
    • $n_2$ = refractive index of the cladding (less dense medium)
Total internal reflection occurs when the angle of incidence exceeds $\theta_c$. The core must have a higher refractive index than the cladding ($n_1 > n_2$).
Power of a lens
$$P = \frac{1}{f}$$
  • Where:
    • $P$ = power of the lens (dioptres, D)
    • $f$ = focal length (m)
Positive for converging lenses, negative for diverging lenses. For lenses in contact, $P_\text{total} = P_1 + P_2 + P_3 + \ldots$
Maximum photon energy (X-ray tube)
$$E_{\max} = eV$$
  • Where:
    • $E_{\max}$ = maximum photon energy (J)
    • $e$ = charge of the electron ($1.6 \times 10^{-19}$ C)
    • $V$ = accelerating voltage across the X-ray tube (V)
The maximum X-ray photon energy equals the full kinetic energy of the accelerated electron. This sets the cut-off wavelength: $\lambda_{\min} = hc / eV$.
Magnification
$$m = \frac{v}{u} = \frac{h_i}{h_o}$$
  • Where:
    • $m$ = magnification (no units)
    • $v$ = image distance (m)
    • $u$ = object distance (m)
    • $h_i$ = image height (m)
    • $h_o$ = object height (m)
$m > 1$ means the image is magnified. $m < 1$ means the image is diminished. A negative $m$ indicates an inverted image.
Acoustic impedance
$$Z = \rho c$$
  • Where:
    • $Z$ = acoustic impedance (kg m$^{-2}$ s$^{-1}$)
    • $\rho$ = density of the medium (kg m$^{-3}$)
    • $c$ = speed of sound in the medium (m s$^{-1}$)
The greater the difference in $Z$ between two media, the more ultrasound is reflected at the boundary. Coupling gel matches the impedance of the transducer to the skin.
Intensity reflection coefficient
$$\alpha = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2}$$
  • Where:
    • $\alpha$ = intensity reflection coefficient (no units, 0 to 1)
    • $Z_1$ = acoustic impedance of medium 1 (kg m$^{-2}$ s$^{-1}$)
    • $Z_2$ = acoustic impedance of medium 2 (kg m$^{-2}$ s$^{-1}$)
The fraction of ultrasound intensity reflected at a boundary. When $Z_1 \approx Z_2$, almost no reflection occurs (good transmission). When $Z_1$ and $Z_2$ are very different (e.g. air and tissue), nearly all intensity is reflected.
Effective half-life
$$\frac{1}{T_E} = \frac{1}{T_P} + \frac{1}{T_B}$$
  • Where:
    • $T_E$ = effective half-life
    • $T_P$ = physical half-life (due to radioactive decay)
    • $T_B$ = biological half-life (due to metabolic removal)
The effective half-life is always the shortest of the three values. It determines how quickly the activity of a tracer inside the body falls to half.
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